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I have convinced myself in a computation-heavy, ad-hoc way that a quadratic extension $K$ of $\mathbb{Q}$ occurs as the unique quadratic subfield of a $\mathbb{Z}/4$-extension of $\mathbb{Q}$ if and only if $N_{K/\mathbb{Q}}(x)=-1$ for some $x\in K$. (See below for proof sketch.) There must be a more conceptually illuminating proof?

Morally / conceptually, why is it that a quadratic field occurs inside a cyclic degree 4 field if and only if -1 occurs in it as a norm?

What's the story here? An optimum answer would be a proof that doesn't ask the reader to follow, trust, or replicate any calculations. (As opposed to the following, where I have written "computation shows" about 5 times.)

Computation-heavy, ad-hoc proof sketch: $\Rightarrow$ If $K=\mathbb{Q}(\alpha)$ with $\alpha^2\in \mathbb{Q}, \alpha\notin\mathbb{Q}$, and $\exists x\in K$ with $N_{K/\mathbb{Q}}(x)=-1$, we have $x=a+b\alpha$, $a,b\in\mathbb{Q}$, with $N(x)=a^2-\alpha^2b^2 = -1$. Let $\beta = \sqrt{\alpha^2b-a\alpha}$ and let $L=K(\beta)$. Then, by computation, $\beta$ satisfies the polynomial $$f=x^4-2\alpha^2bx^2+\alpha^2\in\mathbb{Q}[x]$$ which is irreducible by Eisenstein, thus $\beta^2$ is not a square in $K$, so $\mathbb{Q}(\beta)=L$ is degree $4$. Let $$\beta'=(a+b\alpha)\beta\in L$$ Computation shows $\beta'$ also satisfies $f$, so $\beta\mapsto\beta'$ induces an automorphism of $L$; call it $g$. More computation shows $g(\beta')=-\beta$, so $g$ has order $4$, and $\beta$ has 4 conjugates in $L$, thus $L/\mathbb{Q}$ is Galois, with $\operatorname{Gal}(L/\mathbb{Q})$ generated by $g$. This proves the "if" direction.

$\Leftarrow$ If $L$ is a $\mathbb{Z}/4$ extension of $\mathbb{Q}$, let $g$ be a generator of its Galois group and let $K$ be the unique quadratic subfield. We can take $K=\mathbb{Q}(\alpha)$ with $\alpha^2\in\mathbb{Q}$ and $L=K(\beta)$ with $\beta^2\in K$. $1,\alpha,\beta,\alpha\beta$ is a basis for $L/\mathbb{Q}$; we can write a matrix for $g$, seen as a $\mathbb{Q}$-linear transformation, with respect to this basis. We must have $g(1)=1$, and $g(\alpha)=-\alpha$ because $g$ cannot act trivially on $K$. Let $g(\beta)=c+d\alpha+a\beta+b\alpha\beta$ with $a,b,c,d\in\mathbb{Q}$. Then, using the fact that $g(\alpha\beta)=g(\alpha)g(\beta)$, the matrix of $g$ is

$$\begin{pmatrix}1& &c&-\alpha^2d\\ &-1&d&-c\\ & &a&-\alpha^2b\\ & &b&-a\end{pmatrix}$$

Also, $g^2(\alpha)=\alpha$ because $K$ is fixed by $g^2$, and $g^2(\beta)=-\beta$ because $-\beta$ is $\beta$'s only conjugate over $K$ and $L$ is not fixed by $g^2$. Thus the matrix of $g^2$ is

$$\begin{pmatrix}1& & & \\ &1& & \\ & &-1& \\ & & &-1\end{pmatrix}$$

It follows by computation that $a^2-\alpha^2b^2=-1$. (It also follows that $c=d=0$, but this isn't needed.) Thus $a+b\alpha\in K$ has norm $-1$. This proves the "only if" direction.

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A partial answer (i.e. one direction, and not quite computation-free) is that if $L/\mathbb{Q}$ is a $\mathbb{Z}/4$ extension with Galois group generated by $g$, $K$ is the fixed field of $g^2$, and $\beta$ is a field generator for $L/\mathbb{Q}$ that is the square root of an element of $K$, then $g(\beta)/\beta$ is in $K$ and has norm $-1$. For, as argued in the OP, $g^2(\beta)=-\beta$, thus

$$g^2(g(\beta)/\beta)=g(g^2(\beta))/g^2(\beta)=g(-\beta)/(-\beta) = g(\beta)/\beta$$

so $g(\beta)/\beta\in K$, and its conjugate over $\mathbb{Q}$ must be its image $g^2(\beta)/g(\beta)=-\beta/g(\beta)$ under $g$; thus its $K/\mathbb{Q}$-norm is the product


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I'll refer to my notes for some precise statements and definitions:

Define $I_K^S$ to be the group of ideals in $K$ generated by prime ideals not in $S$ (or not dividing primes in $S$). Define $P_K(1,\mathfrak m)$ to be the group of principal ideals that are "1 modulo $\mathfrak m$." (10.3)

By class field theory (10.4.1), if $L/K$ is abelian, $\mathfrak m$ is a modulus containing the set of primes of $K$ ramifying in $L$, and $S$ the set of primes dividing $\mathfrak m$, there is an isomorphism $\psi_{L/K}:I^S_K/(P_K(1,\mathfrak m)\cdot \text{Nm}_{L/K}(I_L^S))\xrightarrow{\cong} G(L/K)$, and this is compatible with field extension $M/L$ to form a commutative square, the right side of the square being the natural projection $G(M/K)\to G(L/K)$.

The case of $\mathbb Q$:

The above may seem complicated, but it's much easier to understand in the case of $\mathbb Q$, because Kronecker-Weber encapsulates much of the statement above.

Every abelian extension $L/\mathbb Q$ corresponds to $(N,m)$ where $N\subseteq (\mathbb Z/m)^{\times}$ satisfies $(\mathbb Z/m)^{\times}/N\cong G(L/\mathbb Q)$. Indeed, $I_{\mathbb Q}^S/P_{\mathbb Q}(1,m\infty)=(\mathbb Z/m)^{\times}$, and the norm group $\text{Nm}_{L/\mathbb Q}(I_L^S)$ can be thought of as a subgroup of $(\mathbb Z/m)^{\times}$. We can all make this very explicit if we want (and avoid CFT): $L=\mathbb Q(\sum_{j\in N}\zeta_m^j)$, and the isomorphism is given by $j\mapsto (\zeta_m\mapsto \zeta_m^j)$. From this it is clear, for example, that $-1\in N$ iff $L$ is real.

Note that if $m\mid m'$ and $N'$ projects to $N$, then $(N,m)$ and $(N',m')$ correspond to the same extension. For quadratic $L$, the smallest $m$ we can take is the discriminant. (It is the same as the smallest $m$ such that $L\subseteq \mathbb Q(\zeta_m)$.)

The Problem

Let $L/\mathbb Q$ be quadratic. The following are equivalent.

(1) $L/\mathbb Q$ occurs inside a $\mathbb Z/4$ field.

(2) There is $(N,m)$ corresponding to $L$ and $N_1\subseteq N$ such that $(\mathbb Z/m)^{\times}/N_1\cong \mathbb Z/4$.

(3) $L$ is real and the discriminant of $L$ is divisible only by primes $p\equiv 1,2\pmod 4$.

(4) $-1\in \text{Nm}_{L/\mathbb Q}(L)$.

(1)$\iff$(2) follows from the preceding discussion.

(3)$\implies$(2): The argument is elementary but the group theory is messy. Choose $m$ such that $8\mid m$ and the discriminant $d$ divides $m$, and write $m=\prod_i p_i^{e_i}$. Let $h_i:G_i\to \mathbb Z/2$ be the unique nontrivial map. Let $h_i:G_i\to \mathbb Z/2$ be a nontrivial map (the only place we have to be careful is $p_i=2$, here use the map $G_i\cong \mathbb Z/2\oplus (\mathbb Z/2)^{e_i-2}\to (\mathbb Z/2)^{e_i-2}\to \mathbb Z/2$). One can check $N=\ker \sum h_i$ corresponds to the unique real quadratic extension with discriminant $d$. (The image to have in mind is a checkerboard.) Because all $p_i\equiv 1,2\pmod 4$, there exist surjective maps $h_i':G_i\to \mathbb Z/4$ and we can let $N_1=\ker \sum h_i'$. This shows (2).

(3)$\implies$(2): Suppose by way of contradiction $N_1$ works. Write $m=\prod_i p_i^{e_i}$ and keep the notation from above.

First suppose $L$ is not real. Then $-1\not\in N$. But then $-1$ would have order 4 in $G/N_1$, contradiction.

Suppose $k$ is such that $p_k\ne 4k+1,2$. Now $G_k/G_k\cap N\cong \mathbb Z/2$ (as opposed to $\{1\}$) because otherwise $(N',m/p_k)$ would also represent $m$, where $N'$ is the projection of $N$ to $\prod_{i\ne k}G_i$. Now consider $g\in \prod G_i$ such that $g_i=1$ for each $i\ne k$ and $g_k$ is a generator of $G_k$. Then $g\not\in G_k\cap N$ so $g\not\in N$. This means $g$ has order 4 in $G/N_1$. But $\mathbb Z/4$ is not a quotient of $G_k$, contradiction.

For (3)$\iff$(4), note $x^2-my^2$ is solvable over every $\mathbb Q_p$ iff $m$ is only divisible by 2 and $4k+1$ primes. Then use the Hasse norm theorem: for cyclic $L/K$, an element is a global norm iff it is a local norm over every place.

The moral reason why you can detect the possibility of a $\mathbb Z/4$ extension with norms is the correspondence between norm groups and abelian extensions; however, (4) is not immediate from (2) because of the technicality with norms of elements vs. norms of ideals. We've already remarked $-1\in N$ iff $N$ is real, so whether $-1\in N$ only measures part of the failure to be part of a $\mathbb Z/4$-extension.

Possibly there's something more elementary you can do with quadratic forms.

@Ben: $\frac{g(\beta)}{\beta}$ reminds me of cohomology, although I don't know if that has anything to do with it.

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